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Author Topic: attention color whizes: non-typical sRGB/RGB/ProPhoto question  (Read 111657 times)
digitaldog
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« Reply #20 on: January 01, 2011, 04:56:49 PM »
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In the equations sections of Bruce Lindbloom's website he has clearly listed the following matrix for transformation from Adobe RGB (D65) to XYZ


Yes, D65 WP to XYZ. Just as he shows ProPhoto and ColorMatch RGB as D50 to XYZ.

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See the last column, which is [0.1881852 0.0752741 0.9911085], and which are the XYZ for a saturated blue in Adobe RGB with D65 white point, and this is the same set of numbers which I have mentioned above you will get by selecting the D65 in the drop down menu, which Bill did not do in his calculation.


I’m confused. Why would these values have anything to do with a saturated blue?

I’d send Bruce a email, go right to the source. He is quite accessible and will easily clear this up.
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« Reply #21 on: January 01, 2011, 05:03:10 PM »
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I’m confused. Why would these values have anything to do with a saturated blue?


Because, these are the numbers you would get by muliplying that matrix with [0, 0, 1], which is the vector for saturated blue (in any RGB system); [0,0,255] is just a scaling of this basic tristimulus value.

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« Reply #22 on: January 01, 2011, 05:08:50 PM »
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Well again, I’d talk with Bruce, at least in terms of the use of the calculator you are using. The bottom line as far as I can see using any existing set of tools (ColorThink, ColorSync Utility, Gamut Works etc) that allows one to take the existing working space profiles that define ProPhoto RGB and Adobe RGB (1998) do not confirm the idea that It does not seem like that Prophoto RGB is always wider than Adobe RGB. Just the opposite shows true. Do you have any such set of graphing utilities that allow you to map in 2D or ideally in 3D the two profiles?
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« Reply #23 on: January 01, 2011, 05:29:14 PM »
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Do you have any such set of graphing utilities that allow you to map in 2D or ideally in 3D the two profiles?

Well, first you can guage the length of Adobe and Prophoto blue primaries from my diagram below.



Secondly, I don't know how tho_mas produced the following diagram, but isn't that saying the same thing? Though he thoght otherwise. He said that it is showing profiles. But it seems like blue is projecting out of the mesh.  I am no expert in such programs and some of the associated terminology, so may be it is showing something else.

verify what? These things are also available as real profiles in real applications.
This is the shape of the profiles (abscol to D50; Adobe = white)... but this is just the shape of the profiles, totally independed from any real application:


This is what happens in color conversions:


Joofa

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« Reply #24 on: January 01, 2011, 05:51:40 PM »
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I don't know how tho_mas produced the following diagram, but isn't that saying the same thing? Though he thoght otherwise. He said that it is showing profiles. But it seems like blue is projecting out of the mesh.  I am no expert in such programs and some of the associated terminology, so may be it is showing something else.

The are unlabeled so I’m not sure what is what here but yes, the blue primaries would fall outside the CIE chromaticity diagram as its are outside human vision. Yes, it appears these are indeed plots of two profiles (again which profiles are what, I don’t know). But using such a utility and plotting Adobe RGB (1998) and ProPhoto RGB, there is zero question that Adobe RGB (1998) fully fits inside of ProPhoto RGB and that the idea that It does not seem like that Prophoto RGB is always wider than Adobe RGB is not true. Its always a wider gamut working space in all directions.
« Last Edit: January 01, 2011, 05:54:02 PM by digitaldog » Logged

Andrew Rodney
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« Reply #25 on: January 01, 2011, 06:04:03 PM »
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Well, first you can guage the length of Adobe and Prophoto blue primaries from my diagram below.

Just to be perfectly clear, there's nothing sacred about Adobe RGB's "numbers". The funny thing is that it was Mark Hamburg who came up with the numbers while trying to find "RGB Color Spaces" to consider when originally working on the color management of Photoshop 5 (that's PS 5 not CS5 back in 1996/7 I think). He found a color space proposal for I believe an editing space for HDTV that was called SMPTE-240M. Back in Photoshop 5, he included that color space. Only problem is the original proposal that Mark found on the net had a typo on 2 of 3 color coordinates (can't remember which ones). Adobe was contacted after the release of PS 5 and told that A) the color space SMPTE-240M was a "proposed" color space and not yet ratified and B) the original proposal had been amended to correct the two typos and so quit calling it "SMPTE" anything. Adobe chose to rename SMPTE-240M to Adobe RGB (1998).

So, the color coordinates are actually based on typos and it's merely happenstance that the coordinates are those particular numbers.

Kodak had already worked on this problem of developing an RGB working space for digital imaging. They originally released a color space called ROMM-RGB which actually ended up being renamed ProPhoto RGB. ProPhoto RGB predates Adobe RGB (SMPTE-240M) by several years.

In the grand scheme of things, it's mildly interesting to talk about this stuff but the bottom line is that for the purposes of digital imaging for photography, ProPhoto RGB again is the ONLY color space that can contain ALL the colors a camera can captures and ALL the colors recent high-end inkjet printers can print.

Don't let the side discussions confuse the issues...and realize Adobe RGB (1998) is actually an accidental color space brought into being...

I also can tell you a history of sRGB originally developed by Michael Stokes while he worked for HP (I believe). Mike later went to MSFT where sRGB was promoted as the "default" color space for Windows (back in 1997/8). While the color space has always sucked for photography, sRGB was pushed by MSFT because from their point of view, the only color than mattered was the color shown on a computer running Windows. Unfortunately the photographic industry drank MSFT's Kool-Aid and a lot of companies fell for it. Which is one reason a lot of photo labs unfortunately try to tell customers to use sRGB and avoid the whole issue of custom output profiles...

Originally, the 's' in sRGB was thought to mean 'standard' (note the small s). When MSFT started pushing it the 's' seemed to represent 'satanic' RGB...now unfortunately it seems to represent 'stupid' RGB...some people may also consider it to represent 'shitty' RGB but I tend to think of it as interchangeable with 'satanic' and 'stupid' RGB.
« Last Edit: January 01, 2011, 06:10:52 PM by Schewe » Logged
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« Reply #26 on: January 01, 2011, 07:01:12 PM »
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but the bottom line is that for the purposes of digital imaging for photography, ProPhoto RGB again is the ONLY color space that can contain ALL the colors a camera can captures and ALL the colors recent high-end inkjet printers can print.



Jeff, is there not any color space that totally contains the gamut of my Epson 4900, but is smaller than ProPhoto RGB? I just got my new 4900 and was about to look for such a working color space. Indeed, I do see where the gamut of my 4900 is clipped in the blue green area by BetaRGB.

Thanks,

Jeff
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« Reply #27 on: January 01, 2011, 07:27:23 PM »
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The are unlabeled so I’m not sure what is what here but yes, the blue primaries would fall outside the CIE chromaticity diagram as its are outside human vision. Yes, it appears these are indeed plots of two profiles (again which profiles are what, I don’t know).

My guess is that thomas' top image showing blue ripping through mesh corresponds to the case that I have labelled as (1) in my note on dpreview (Adobe=D65, Prophoto=D50). And, his bottom image corresponds to my cases (2) or (3), i.e., (Adobe=D65, Prophoto=D65)  and (Adobe=D50, Prophoto=D50), respectively. But, I'm no expert in such prsentations so don't count on me.

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But using such a utility and plotting Adobe RGB (1998) and ProPhoto RGB, there is zero question that Adobe RGB (1998) fully fits inside of ProPhoto RGB and that the idea that It does not seem like that Prophoto RGB is always wider than Adobe RGB is not true. Its always a wider gamut working space in all directions.

After presenting all the numbers, diagrams and discussion, I dont' know how to convince you.

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« Reply #28 on: January 01, 2011, 07:42:06 PM »
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In the grand scheme of things, it's mildly interesting to talk about this stuff but the bottom line is that for the purposes of digital imaging for photography, ProPhoto RGB again is the ONLY color space that can contain ALL the colors a camera can captures and ALL the colors recent high-end inkjet printers can print.

Firstly, thanks for the historical perspective in your insightful note. It was instructive.

I have said before that I can agree with you that it might be a little academic arguing about some blues that can't be observed. However, based upon the numbers Bill produced, I have an unsettling feeling that it might appear that Adobe has implemented their own space, Adobe RGB, with D50, instead of the standard specification of D65 white point, in Photoshop! I think it is a blasphemy to mention that. But, can that happen? Huh  Huh Huh Huh

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« Reply #29 on: January 01, 2011, 07:46:23 PM »
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My guess is that thomas' top image showing blue ripping through mesh corresponds to the case that I have labelled as (1) in my note on dpreview (Adobe=D65, Prophoto=D50). And, his bottom image corresponds to my cases (2) or (3), i.e., (Adobe=D65, Prophoto=D65)  and (Adobe=D50, Prophoto=D50), respectively.

There is NO such thing as ProPhoto RGB working space with a D65 WP. There is no such thing as Adobe RGB (1998) working space with D50 WP. An RGB working space is by its very definition and design, specific values that define its WP, chromaticity values and gamma (TRC). Open Photoshop, go into the Color Settings. Select the RGB working space (like ProPhoto RGB), then toggle to Custom RGB... in the popup menu. See the resulting dialog? Its the DNA of the working space. Alter any value, its not that working space any more.

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After presenting all the numbers, diagrams and discussion, I dont' know how to convince you.

After viewing the 3D gamuts of each, I’d agree.
« Last Edit: January 01, 2011, 07:48:47 PM by digitaldog » Logged

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« Reply #30 on: January 01, 2011, 07:55:51 PM »
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There is NO such thing as ProPhoto RGB working space with a D65 WP. An RGB working space is by its very definition and design, specific values that define its WP, chromaticity values and gamma (TRC).

You got to be kidding. A 3D color space has already three independent vectors, R, G, and B. The standard definition of dimensionality of a finite dimension space is that any 4th vector is linearily dependent on the three. Then why is a white point (4th vector) specified? Think about it.

It is perfectly feasible to conceive an Adobe RGB space with D50 white point instead of the standarized D65 and Prophoto RGB space with a white point of D65 instead of standarized D50.

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After viewing the 3D gamuts of each, I’d agree.

The graph in one of previous messages that I quote above, which shows Adobe and ProPhoto RGB primaries, gives an indication of the gamut. Just that it is not presented in the way you are used to seeing in other programs. But trust me it shows that information if you want to have a ball-park estimate of it. On my computer I can rotate it around and have a better view of certain axis. But I selected that view because it shows the angular relationship of the 6 primaries in addition to their magnitudes. You can connect the endpoints of the vectors displayed and have a triangle and project the spectral colors that I have also drawn, and hey, you have your chromacity diagram for the Adobe and ProPhoto Systems.

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« Reply #31 on: January 01, 2011, 08:21:55 PM »
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I have said before that I can agree with you that it might be a little academic arguing about some blues that can't be observed. However, based upon the numbers Bill produced, I have an unsettling feeling that it might appear that Adobe has implemented their own space, Adobe RGB, with D50, instead of the standard specification of D65 white point, in Photoshop! I think it is a blasphemy to mention that. But, can that happen? Huh  Huh Huh Huh
Joofa

Joofa,

Not quite so academic, since there is no RGB value in the AdobeRGB gamut that can not be seen by the human eye, since the AdobeRGB coding efficiency (as determined by Bruce Lindbloom) is 100%. On the other-hand, only 50% of colors (the CIE L*a*b gamut) are encoded by AdobeRGB.  The critical gamut for most photography is that of real world surface colors: colors that exist in nature and are seen via reflected light. Gernot Hoffman has some plots showing its gamut, which is larger than AdobeRGB but less than ProPhotoRGB. Neon lights and other emissive sources exceed this gamut. I think that Bruce Lindbloom tried to approximate the real world surface colors in his BetaRGB.

Regards,

Bill
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« Reply #32 on: January 01, 2011, 09:53:55 PM »
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Joofa,

Your XYZ graph doesn't show the relative gamuts of the two spaces. Just because one blue vector has a higher magnitude than the other isn't enough. The direction is important too. You have only proven that the ARGB blue primary is further from the origin than the prophoto RGB primary. That doesn't mean what you think it means.

Bruce has a 3D gamut utility on the his website here: http://www.brucelindbloom.com/index.html?Eqn_RGB_XYZ_Matrix.html (scroll down)

It will show the AdobeRGB space completely surrounded by the prophoto space. For me this is definitive, but it doesn't seem to convince you so lets try a little math.

Let's compare pure blue in both spaces. Take your AdobeRGB blue [0, 0, 1.0] and convert it to XYZ you also have to account for the chromatic adaptation since we have two different white points. Bruce makes it easy to do by providing a adobeRGB matrix with an adaptation to d50 built in:


[0.6097559 0.2052401 0.1492240]
[0.3111242 0.6256560 0.0632197]   x [0 0 1.0] = [ 0.149224   0.0632197  0.7448387]
[0.0194811 0.0608902 0.7448387]

This is AdobeRGB's pure blue converted to XYZ and D50.

Now lets convert it to ProPhotoRGB:
[ 1.34594337 -0.25560752 -0.05111183]
[-0.54459882  1.5081673    0.02053511] x [0.149224   0.0632197  0.7448387] = [ 0.14661755  0.02937401  0.90260503]
[ 0.000       0.000       1.21181275]

Apply gamma correction:
[ 0.14661755  0.02937401  0.90260503] ^ (1/1.8) = [ 0.3441676   0.14088621  0.94466218]

[ 0.3441676   0.14088621  0.94466218] is how you represent adobeRGB's pure blue in prophotoRGB with room to spare.

If you multiply by 255 you'll get:
[  87.76273719   35.92598274  240.88885588] which is pretty much what photoshop will tell you if you do the profile conversion there.

Now try to do the opposite. Start with ProPhotoRGB's [0, 0, 1] convert to XYZ and then try going to AdobeRGB. You will get negative RGB numbers showing that this color is out of gamut.

Of course this could all be proven wrong if you show me an AdobeRGB color that can't be represented in prophoto space.
« Last Edit: January 01, 2011, 10:23:50 PM by MarkM » Logged

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« Reply #33 on: January 01, 2011, 10:28:54 PM »
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Jeff, is there not any color space that totally contains the gamut of my Epson 4900, but is smaller than ProPhoto RGB?

I don't know...all I know for a fact is there is no color space (that I know of) that can contain ALL the colors my cameras can capture and my printer can print other than ProPhoto RGB...pretty sure Beta RGB and Ekta RGB (from Joe Homes) can't say that.

So, in the grand scheme of things, I really don't care.

To the best of my knowledge I've NEVER had any image quality issues caused by using ProPhoto RGB in 16 bit/channel. Yes, it's a huge color space but I've never seen any real problems from using that space. I HAVE seen real problems using Adobe RGB and sRGB. Since these three color spaces are the only color spaces I've used on a regular basis and are one of 4 offered in Camera Raw, the only choice for me is ProPhoto RGB. And I've talked at length to the likes of Thomas Knoll, Mark Hamburg, Michael Stokes and Karl Lang (to name a few) and they all pretty much point to ProPhoto RGB as being the best large gamut color space to using for digital imaging. Which is why ProPhoto RGB with a linear gamma is the internal working space of both Camera Raw and Lightroom.

I used to use ColorMatch RGB which was originally developed by Karl Lang when he worked for Radius and did the original PressMatch displays but it suffers from a deficient in cyan when converting to CMYK.

I know some people advocate using the largest, most efficient color space that can handle input and output but the only one that I've seen that never clips any colors is Pro Photo RGB.

And I simply don't care about the fact that theoretically, Pro Photo RGB is "inefficient". I've never had any problems because Pro Photo RGB is inefficient and "too big". Yes, it's a large color space and that means that the massive separation of levels due to the color coordinates "could" cause efficiency problems-it's just I've never seen any actual proof of that and I've never seen any problems with my own work.

Use whatever color space makes you happy...I use only ProPhoto RGB for my own work, I use sRGB for the web and I use Adobe RGB when sending images to anybody who I don't know for an absolute fact know what they are doing which includes most all magazines, ad agencies, photo labs and designers...(which basically means everybody else in the world).

:~)
« Last Edit: January 01, 2011, 10:31:48 PM by Schewe » Logged
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« Reply #34 on: January 01, 2011, 11:34:32 PM »
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Joofa,

Your XYZ graph doesn't show the relative gamuts of the two spaces. Just because one blue vector has a higher magnitude than the other isn't enough. The direction is important too. You have only proven that the ARGB blue primary is further from the origin than the prophoto RGB primary. That doesn't mean what you think it means.

Please don't assume that what I think I mean. Do you really think that I plotted the angles between various primaries just arbitrarily?

Quote
Bruce has a 3D gamut utility on the his website here: http://www.brucelindbloom.com/index.html?Eqn_RGB_XYZ_Matrix.html (scroll down)

It will show the AdobeRGB space completely surrounded by the prophoto space. For me this is definitive, but it doesn't seem to convince you so lets try a little math.

Let's compare pure blue in both spaces. Take your AdobeRGB blue [0, 0, 1.0] and convert it to XYZ you also have to account for the chromatic adaptation since we have two different white points. Bruce makes it easy to do by providing a adobeRGB matrix with an adaptation to d50 built in:


[0.6097559 0.2052401 0.1492240]
[0.3111242 0.6256560 0.0632197]   x [0 0 1.0] = [ 0.149224   0.0632197  0.7448387]
[0.0194811 0.0608902 0.7448387]


I believe this is the incorrect matrix for this situation. This is for conversion from Adobe RGB with D50 white point to XYZ. I mentioned before that the Bruce has also provided the matrix for Adobe RGB with D65 white point conversion to XYZ, and I repeat it again here:

0.5767309  0.1855540  0.1881852
0.2973769  0.6273491  0.0752741
0.0270343  0.0706872  0.9911085

[Rest of calculation snipped.]

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This is AdobeRGB's pure blue converted to XYZ and D50.

What pure blue? The "pure" blue in Adobe RGB with D65 is [0.1881852 0.0752741 0.9911085] and its representation in Adobe RGB with D50 white point is *not* [0 0 1]! The representation of [0 0 1] in Adobe RGB with D50 white point with exact calculation is [0.137826   0.055130   0.725885], but  you can use the one you used in your matrix of [0.149224   0.0632197  0.7448387] also as it is close enough.

I mentioned before and I repeat it again here the colors represented by [0 0 1] in Adobe RGB with D65 and D50 are different. The one with D50 can be represented in Prophoto RGB with D50 white point, but not the one with D65.

Joofa
« Last Edit: January 09, 2011, 06:24:13 AM by joofa » Logged

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« Reply #35 on: January 02, 2011, 12:18:50 AM »
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This is for conversion from Adobe RGB with D50 white point to XYZ

No, it's not. As others keep pointing out, there is no Adobe RGB with a D50 white point.

If you want to compare spaces with different white points, you need to convert to a common white point. To go from AdobeRGB to ProPhotoRGB you need to account for the different white points. The AdobeRGB D50 matrix is just a convenience to remove that step from the calculations. You can do the same thing using the D65 matrix, but then you need to explicitly perform the chromatic adaptation yourself.

Like this:

AdobeRGB -> XYZ using the D50 matrix (what I originally did):
[0.6097559 0.2052401 0.1492240]
[0.3111242 0.6256560 0.0632197]   x [0 0 1.0] = [ 0.149224   0.0632197  0.7448387]
[0.0194811 0.0608902 0.7448387]

or like this:
AdobeRGB -> XYZ using the D65 matrix:
[0.5767309 0.1855540 0.1881852]
[0.2973769 0.6273491 0.0752741] x [0 0 1.0] = [ 0.1881852  0.0752741  0.9911085]
[0.0270343 0.0706872 0.9911085]

But now you need to explicitly account for D50 if you want to move to a D50 space or compare to a D50 space (using Bradford):
[1.0478112 0.0228866 -0.0501270]
[0.0295424 0.9904844 -0.0170491] x [ 0.1881852  0.0752741  0.9911085] = [ 0.14922403  0.06321976  0.74483862]
[-0.0092345 0.0150436 0.7521316]

See, same result. This is all the AdobeRGB D50 matrix is doing. Again there is no AdobeRGB D50. Making a statement like "[0 0 1] in Adobe RGB with D50" doesn't make any sense. You can convert to XYZ and then to d50 which is what I did above, but if you want to get back to Adobe RGB you need to convert back to D65.
« Last Edit: January 02, 2011, 12:53:36 AM by MarkM » Logged

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« Reply #36 on: January 02, 2011, 12:49:47 AM »
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You got to be kidding. A 3D color space has already three independent vectors, R, G, and B. The standard definition of dimensionality of a finite dimension space is that any 4th vector is linearily dependent on the three. Then why is a white point (4th vector) specified? Think about it.

The point is, white point isn't a vector. It's a point.  Grin

But in all seriousness, Andrew is entirely correct. This discussion is the color space equivalent of "if my aunt was a man, she'd be my uncle". ProPhoto is a D50 color space. You can construct a infinite number of possible alternate color spaces, e.g., one with a D65 white point and ProPhoto primaries. But to suggest that the ProPhoto color space doesn't contain a specific color because your new alternate color space with ProPhoto primaries doesn't contain that point is incorrect. The new space that you invented may have the same primaries as ProPhoto, but it isn't ProPhoto. Color space definitions have white points for a reason.

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« Reply #37 on: January 02, 2011, 01:09:02 AM »
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No, it's not.

Yes it is.

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As others keep pointing out, there is no Adobe RGB with a D50 white point.

Selection of a white point is independent of the selection of the primaries. Adobe RGB primaries with the same chromacity coordinates can be used with either D65 white point or D50 white point. It is a vector space. You are given three basis vectors (RGB) directions (chromacity coordinates) but you don't know how long is a "unit vector". The white point sets the units. This concept is an important one and apparently is being missed by you and others.

With this understanding Adobe RGB with D65 and D50 are two different coordinate systems in the same 3D space.

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If you want to compare spaces with different white points, you need to convert to a common white point.
To go from AdobeRGB to ProPhotoRGB you need to account for the different white points. The AdobeRGB D50 matrix is just a convenience to remove that step from the calculations. You can do the same thing using the D65 matrix, but then you need to explicitly perform the chromatic adaptation yourself.

Like this:

AdobeRGB -> XYZ using the D50 matrix (what I originally did):
[0.6097559 0.2052401 0.1492240]
[0.3111242 0.6256560 0.0632197]   x [0 0 1.0] = [ 0.149224   0.0632197  0.7448387]
[0.0194811 0.0608902 0.7448387]

AdobeRGB -> XYZ using the D65 matrix:
[0.5767309 0.1855540 0.1881852]
[0.2973769 0.6273491 0.0752741] x [0 0 1.0] = [ 0.1881852  0.0752741  0.9911085]
[0.0270343 0.0706872 0.9911085]

But now you need to explicitly account for D50 if you want to move to a D50 space or compare to a D50 space (using Bradford):
[1.0478112 0.0228866 -0.0501270]
[0.0295424 0.9904844 -0.0170491] x [ 0.1881852  0.0752741  0.9911085] = [ 0.14922403  0.06321976  0.74483862]
[-0.0092345 0.0150436 0.7521316]

See, same result. This is all the AdobeRGB D50 matrix is doing.

What all the above calculation is doing is that showing that for e.g., if you take a 100% reflector and shine D65 light on it and measure the XYZ tristimulus of blue in an rgb mixture to match D65, you will get blue = [0.1881852  0.0752741  0.9911085]. If instead of D65 you have shone D50 on the same reflector you would have measured [0.14922403  0.06321976  0.74483862]. But these two colors are different in absolute terms. What Bradford transformation says is that you don't need to shine a D50 light and measure the XYZ. If you have D65 tristimulus then you can convert from D65 to D50 with certain human assumptions in mind regarding neutral/gray colors consistency.

And this is what you have shown. But this is not what I'm after.

Instead of a reflector, suppose there is a source that emits two colors A = [0.1881852  0.0752741  0.9911085] and B= [0.14922403  0.06321976  0.74483862] in XYZ, which are, in absolute terms, two different colors. If you measure A with Adobe RGB primaries scaled to D65 white point you get [0, 0, 1]. If you measure B with Adobe RGB primaries scaled to D50 white points you get [0, 0, 1]. But what if you measure A with D50 and B with D65. You, of course, don't get [0, 0, 1], but some other numbers, which can be calculated, but not important right now.

Now lets try to measure A and B in Prophoto RGB scaled to D50 white point. You will find that A needs more than unit amount of blue Prophoto RGB while B needs less. Which means that A can't be represented but B can be in Prophoto (D50). But A can be represented in Adobe RGB (D65) as [0,0,1]. So this is a color which has representation in Adobe RGB (D65) but not in Prophoto RGB (D50) without clipping.

There is no need for a Bradford transformation here as we are doing a direct measurement of two different colors A and B in a measurement system using ProPhoto RGB scaled to D50 white point.

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Again there is no AdobeRGB D50.

I hope by now you know that one can be constructed as easily as one with D65!

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Making a statement like "[0 0 1] in Adobe RGB with D50" doesn't make any sense.

Again, I hope by now you understand.

Sincerely,

Joofa
« Last Edit: January 10, 2011, 11:39:51 AM by joofa » Logged

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« Reply #38 on: January 02, 2011, 01:14:12 AM »
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The point is, white point isn't a vector. It's a point.  Grin

Yeah right. You have a good grasp of colorimetry!

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But to suggest that the ProPhoto color space doesn't contain a specific color because your new alternate color space with ProPhoto primaries doesn't contain that point is incorrect. The new space that you invented may have the same primaries as ProPhoto, but it isn't ProPhoto. Color space definitions have white points for a reason.

I don't think you have understood what I'm saying here. I'm not inventing new color spaces. The basic premise of my note is the standard Adobe RGB (D65) and standard ProPhoto RGB (D50). Other spaces with the same primaries but different white points were to illustrate what is going on here.

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« Reply #39 on: January 02, 2011, 03:01:17 AM »
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Selection of a white point is independent of the selection of the primaries
Right, and you can select the blue primary independently from the red. When you do that you create a new colorspace.

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With this understanding Adobe RGB with D65 and D50 are two different coordinate systems in the same 3D space

If they are two different coordinate systems, they are two different color spaces. One, the one with the D65 white point is called AdobeRGB 1998, you can read it's spec here: http://www.adobe.com/digitalimag/pdfs/AdobeRGB1998.pdf

In that document you will learn the chromatically coordinates of the primaries and the white point (x=0.3127, y=0.3290, which corresponds to D65).

The other space you describe whose primaries have the same chromaticity coordinates as AdobeRGB but with a different white point, let's call it joofaSpace, is something else. I don't know much about joofaSpace, but I do know that by definition it is not AdobeRGB. If you still insist this is not true, you need to find a spec for adobe RGB that doesn't include a d65 white point.

Another thing, in all this discussion we have not seen a transformation matrix from joofaSpace to XYZ. The matrix you would like to use—the D50 matrix we've been using above, does not actually share the primaries from AdobeRGB. You can very easily calculate the chromaticity coordinate of the primaries from the matrix. If you do this for the AdobeRGB D65 matrix you get these coordinates (exactly what they should be):
Red:  0.6400 0.3300
Green:  0.2100 0.7100
Blue:  0.1500 0.0600

If you do it for the D50 matrix you get:
Red:  0.6484 0.3309
Green:  0.2301 0.7016
Blue:  0.1559 0.0660

They're significantly different. So if your goal is to use a transformation matrix with the same primaries as AdobeRGB but with a different white point, you're using the wrong one. Like I posted above, that isn't what that matrix does.

If you'd like a matrix for joofaSpace, I've calculated one for you:
[.645 .181 .138]
[.333 .612 .055]
[.030 .069 .726]

I did it by hand so I only calculated three digits, but if you run the numbers you'll find that it delivers a white point that corresponds to D50 and the primaries match those of AdobeRGB. Just don't call it AdobeRGB 1998.
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